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[打法交流]
近期在网上研究到了个百家乐著名战术 蒙特卡洛战术
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百家乐蒙特卡洛战术，需要纸和笔。但是不很难。
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" v4 i' Z9 G% {/ E(1) 首先写「 1 2 3 」的数列。
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(2) 赌本就是左端的数字 (1) 加右端的数字 (3) 即是 (4) 。# q# x* Z) J& P# k+ ?
7 U4 U7 l; ]7 o4 a& Y(3) 输了的话就在数列右端之外又加上顺序的下一个数，现在这个例子要加的数字 (=4) ->( 「 1 2 34 」 ) 。如此类推，并且重複上面的第 (2) 点 ( 但今次会是 (1)+(4)=5) 。
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# c) E" {' F0 @5 J4 f(4) 胜了的话，倍率 2 倍的游戏就把左端的数字 (1) 和右端的数字 (3) 消去。倍率 3 倍的游戏从数列上去掉左端的二个数字和右端的二个数字。
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(5) 如果数列所有的数字都消了就代表结束。
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(6) 想继续的话，回到 (1) 。
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! f4 W$ f! c& j6 t! ?6 b' Y蒙特卡洛战术的问题，在于如果没有充分而源源不绝的赌金，就会出现途中被迫停止的可能。而且连败的话，亦有机会超越 Max Bet ( 赌注的上限额 ) 。
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二合一战术
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这个可说是蒙特卡洛战术的一种，是以 1 局取回 2 局所输的方法。方法简单。写下输了的金额，赌注就那个左端和右端的数字的共计金额。1 Q6 r" Y' Z0 _& @8 p0 A5 T
8 {" G* ^ Q/ L3 E+ l9 L" q$ S譬如，先写下「 1 」。所以第一局的赌注是 1 。如这局输了，右侧添写 1 ，成为「 1 1 」。那个下一局赌注将会是左端和右端的共和 =2 。
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反复这样做，如四连败，数列将成为「 1 1 2 3 」。因此下一赌注会是 4 ，而且胜了的话，消去左端和右端的数字，数列将成「 1 2 」，下一赌注就等于 1 和 2 的总数 3 。再胜的话，会变成？% U, ]9 b; |0 l' K/ ]0 K7 z
) q# \6 L" W8 G( E儘管结果是 2 胜 4 败，共计收支是一加一减等于零。
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这方法我试了不咋地，你们可能可以?
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